Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{r^2 - 7r}{r - 9} \times \dfrac{r^2 + r - 90}{r^3 - 3r^2 - 28r} $
Solution: First factor out any common factors. $z = \dfrac{r(r - 7)}{r - 9} \times \dfrac{r^2 + r - 90}{r(r^2 - 3r - 28)} $ Then factor the quadratic expressions. $z = \dfrac {r(r - 7)} {r - 9} \times \dfrac {(r - 9)(r + 10)} {r(r - 7)(r + 4)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {r(r - 7) \times (r - 9)(r + 10) } {(r - 9) \times r(r - 7)(r + 4) } $ $z = \dfrac {r(r - 9)(r + 10)(r - 7)} {r(r - 7)(r + 4)(r - 9)} $ Notice that $(r - 7)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {r(r - 9)(r + 10)\cancel{(r - 7)}} {r\cancel{(r - 7)}(r + 4)(r - 9)} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $z = \dfrac {r\cancel{(r - 9)}(r + 10)\cancel{(r - 7)}} {r\cancel{(r - 7)}(r + 4)\cancel{(r - 9)}} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $z = \dfrac {r(r + 10)} {r(r + 4)} $ $ z = \dfrac{r + 10}{r + 4}; r \neq 7; r \neq 9 $